If is a banded matrix with a banded inverse then BC F is a product of blockdiagonal matrices We review this fact or ization in which the are tridiagonal and is independent of the matrix size For a permutation with bandwidth each exchanges disjoi ID: 60068
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BandedMatriceswithBandedInversesandA=LPUGilbertStrangAbstract.IfAisabandedmatrixwithabandedinverse,thenA=BC=F1:::FNisaproductofblock-diagonalmatrices.Wereviewthisfactor-ization,inwhichtheFiaretridiagonalandNisindependentofthematrixsize.Forapermutationwithbandwidthw,eachFiexchangesdisjointpairsofneighborsandN2w.Thispaperbeginstheextensiontoinnitematrices.Fordoublyinnitepermutations,thefactorsFnowincludetheleftandrightshift.Forbandedinnitematrices,wediscussthetriangularfactorizationA=LPU(completedinalaterpaperonTheAlgebraofElimination).Fourdirectionsforelimina-tiongivefourfactorizationsLPUandUPLandU1U2(Bruhat)andL1L2withdierentL,U,Pand.1.IntroductionThispaperisabouttwofactorizationsofinvertiblematrices.OneisthefamiliarA=LU,isthelowertimesuppertriangular,whichisacompactdescriptionoftheeliminationalgorithm.ApermutationmatrixPmaybeneededtoexchangerows.ThequestioniswhetherPcomesbeforeLorafter!NumericalanalystsputPrst,toordertherowssothatallupperleftprincipalsubmatricesbecomenonsingular(whichallowsLU).AlgebraistswriteA=LPU,andinthisformPisunique.Mostmathematiciansthinkonlyofoneortheother,andasmallpurposeofthispaperistopresentboth.Wealsoconnecteliminationstartingatthe(n;1)entrytotheBruhatfactorizationA=U1U2.Inthisformthemostlikelypermutation(betweentwouppertriangularfactors)isthereverseidentity.Infactthefournaturalstartingpoints(1;1),(n;1),(n;n),(1;n)leadtofourfactorizationswithdierentL,U,P,:A=LPU;A=UU;A=UPL;A=LL:TheP'sand'sareuniquewhenAisinvertible,andineachcaseeliminationcanchooseroworcolumnoperationstoproducethesefactorizations.Ourlargerpurposeistodiscussbandedmatricesthathavebandedinverses:Aij=0andalso(A 1)ij=0forji jj-1.6;鑖w.(TheuniquepermutationPwillsharethesamebandwidthw.)ThepurposeofourfactorizationA=F1:::FNistomakethisbandednessevident:ThematricesFiareblockdiagonal.Then AMSSubjectClassication.PROC10959.Keywordsandphrases.Bandedmatrix,Bandedinverse,Bruhatpermutation,Factorization,Groupgenerators,Shiftingindex.1 2GILBERTSTRANGtheirinversessinglyarealsoblockdiagonal,andtheproductsA=F1:::FNandA 1=F 1N:::F 11arebothbanded.WeestablishedthisfactorizationinTheAlgebraofElimination[16]usingAs-plund'stestforabandedinverse:AllsubmatricesofAabovesubdiagonalworbelowsuperdiagonalwhaverankw.ThekeypointofthetheoremisthatthenumberNCw2offactorsFiiscontrolledbywandnotbythematrixsizen.Thisopensthepossibilityofinnitematrices(singlyordoublyinnite).Wewillnotachieveherethecompletefactorizationsofinnitematrices,butwedodescribeprogress(aswellasdiculties)forA=LPU.Inoneimportantcase|bandedpermutationsofZ,representedbydoublyinnitematrices|weintroduceanideathatmaybefruitful.ThefactorsF1;:::;FNforthesematricesincludedisjointtranspositionsTofneighborsandalsobi-inniteshiftsSandST.Allhavew=1:T=26666401100110377775andS=2666400100137775GretaPanovausedaneatvariation[11]ofthe\wiringdiagram"forP,toshowthatthenumberoffactorsisN2w 1.Thisconjecturefrom[16]wasfornitematrices.TheextensiontobandedpermutationsofZallowsalsosshiftfactors.Wecalls(P)theshiftingindexofP(negativeforST=S 1).(Importantandrecentlydiscoveredreferencesare[9;14;15],pleaseseebelow.)Thisshiftingindexhasthepropertythat(1.1)s(P1P2)=s(P1)+s(P2):ItshouldalsohaveausefulmeaningforA,whenA=LPU.FortheperiodicblockToeplitzcasewithblocksizeB,s(P)isthesumofkiintheclassicalfactorizationofamatrixpolynomialintoa(z)=`(z)p(z)u(z)withp(z)=diag(zk1;:::;zkB).Theoriginalpaper[16]outlinedalgorithmstoproducetheblockdiagonalfac-torsFiinparticularcaseswithbandedinverses:1.WaveletmatricesareblockToeplitz(periodic)anddoublyinnite(i;jinZ).Atypicalpairofrowscontains2by2blocksM0toMN 1.TheactionofthisAisgovernedbythematrixpolynomialM(z)=PMjzj.TheinverseisbandedexactlywhendetM(z)hasonlyonetermczN 1.Inthenondegeneratecase,thenumberNcountsthefactorsFiandalsoequalsthebandwidthw(aftercentering):A=2664M0:::MN 1M0:::MN 13775hasfactorsFi=2664BiBi3775:Importantpoint:The2by2blocksBi+1inFi+1areshiftedbyonerowandcolumnrelativetoBiinFi.OtherwisetheproductofF'swouldonlybeblockdiagonal.2.CMVmatrices.Thematricesstudiedin[3;10]havetwoblocksoneachpairofrowsofA.Those2by2blocksaresingular,createdbymultiplyingatypical BANDEDMATRICESWITHBANDEDINVERSESANDA=LPU3F1F2=A(noticeagaintheshiftinpositionoftheblocks):(1.2) 26666641234567893777775266664abcdefghi377775=2666664ab2c2d3e3f4c4d5e5f6g6h7i8g8h9i3777775 Ahasbandwidthw=2.AlsoA 1=F 12F 11hasw=2.Notehowcolumn2ofF1timesrow2ofF2producesthesingularblockwith2c,2d,4c,4d(andtwootherblocksarealsosingular).NecessarilyApassesAsplund'stest:Thosethreesingularblocksassurethateveryadmissiblesubmatrixhasrank2.Onesubmatrixisindicated,abovethesecondsubdiagonalofA.Inapplicationstoorthogonalpolynomialsonthecirclejzj=1,CMVmatricesarenolongerblockToeplitz.The2by2blocksarealldierentasshown.Wemaythinkofthemas\time-varying"waveletmatrices.Extendingthisanalogy,weallowthemtohaveN2blockscenteredalongeachpairofrows(thenw=N).ThefactorizationofAisrecursive.AlwaysFi+1hasitsdiagonalblocksshiftedwithrespecttoFi,asinthemultiplicationabove.3.Orthogonalmatrices.TheoriginalCMVmatricesandtheDaubechieswaveletmatriceswerebandedandalsoorthogonal:ATA=I.Itisnaturaltolookforor-thogonalfactorsFiwithw=1.Thiscanbeachievedforallbandedorthogonalmatrices.Ourexample[10]isaCMVmatrixandDaubechiesmatrixofparticularinterest:F1andF2areToeplitz(periodic)andtheirblocksarerotations:26641+p 3 1+p 31 p 31+p 337752664p 3 11p 3p 3 11p 33775=26641+p 33+p 33 p 31 p 31 p 3 3+p 33+p 3 1 p 33775(1.3)Tonormalizethosecolumnstounitlength,divideby2p 2,2,and4p 2.Therotationanglesare=12and=6,addingto=4.4.Permutationmatrices.Theordering(3;4;1;2)isassociatedwitha4by4permutationmatrix.ThebandwidthsofPandPT=P 1arew=2.Thisisthegreatestdistancew=maxji p(i)jthatanyentrymustmove:(3;4;1;2)correspondstoP=266400100001100001003775:Three\greedysteps"[1]willexchangedisjointneighborstoreachtheorder(1;2;3;4):(3;4;1;2)!(3;1;4;2)!(1;3;2;4)!(1;2;3;4): 4GILBERTSTRANGTheproductofthecorrespondingblockdiagonalmatricesF1F2F3isP:(1.4)P=26641011013775266401100110377526641011013775:InthisexampleNreachesitsmaximumvalue2w 1=3forpermutationsofbandwidthw.2.TheFactorizationA=F1:::FNAandA 1arebandednbynmatrices.Wewilldisplaythestepsoftheirfactorizationintoblockdiagonalmatrices.Thefactorsarereachedintwosteps:(1)FactorAintoBCwithdiagonalblocksofsizesw;2w;2w;:::forBand2w;2w;:::forC.Asinequations(2),(3),(4),thisshiftbetweenthetwosetsofblocksmeansthatA=BCneednotbeblockdiagonal.(2)BreakBandCseparatelyintofactorsFwithblocksofsize2(or1)alongthediagonal.Thisisachievedin[17]byordinaryelimination,andisnotrepeatedhere.InprinciplewemayneedO(w2)steps,movingupwardinsuccessivecolumns1;:::;2wofeachblockinBandC.WedowanttoexplainthekeyideabehindStep1,toreachA=BC.SupposeAhasbandwidthw=2.IfA 1alsohasw=2,Asplund'stheorem[2;18]imposesarankconditiononcertainsubmatricesofA,abovesubdiagonalworbelowsuperdiagonalw.Allthesesubmatricesmusthaverankw.ToapplythisconditionwetaketherowsandthecolumnsofAingroupsofsize2w=4.ThemaindiagonalisindicatedbycapitallettersX,andAsplund'sconditionrank2appliestoallHiandKi.Allthoseranksareexactly2becauseeachsetoffourrowshasrank4.H1Xx xK1xX xxxx Xxxx xXxx H2xxXx xK2xxX xxxx Xxxx xXxx OurplanistodiagonalizethesesubmatricesH1;K1;H2;K2;:::byrowoperationsontheH'sandcolumnoperationsontheK's.TherowstepscanbedoneinparallelonH1;H2;:::togivetheblocksinB.ThecolumnstepsgivetheblocksinC,andwefoldintoCthediagonalmatrixreachedattheendoftheelimination.EliminationonH1:Withrank2,rowoperationswillreplaceeveryxbyzero.Rows3and4ofthenewK1mustnowbeindependent(sincetheyhaveonlyzerosinH1).EliminationonK1:Withrank2,upwardrowoperationsandthencolumnoperationswillreplaceeveryxbyzero.Columns5and6inthenewH2mustnowbeindependent(sincethosecolumnsstartwithzerosinthecurrentK1). BANDEDMATRICESWITHBANDEDINVERSESANDA=LPU5EliminationonH2:Leftwardcolumnoperationsandthenrowoperationswillreplaceeveryxbyzero.EventuallythewholeAisreducedtoadiagonalmatrix.3.FourTriangularFactorizationsThebasicfactorizationisA=LU.Therstfactorhas1'sonthediagonal,thesecondfactorhasnonzerosd1;:::;dn.Multiplyingthekbykupperleftsub-matricesgivesAk=LkUk,soanecessaryconditionforA=LUisthateveryAkisnonsingular.Executingthestepsofeliminationshowsthatthisconditionisalsosucient.Afterk 1columnsarezerobelowthemaindiagonal,the(k;k)entrywillbedetAk=detAk 1.Belowthisnonzeropivotdk,rowoperationswillachievezerosincolumnk.Thencontinuetok+1.InvertingallthoserowoperationsbyLleavesA=LU.Notethatcolumnoperationswillgiveexactlythesameresult.Atstepk,subtractmultiplesofcolumnkfromlatercolumnstoclearoutrowkabovethediagonal.AfternstepswehavealowertriangularLcwiththesamepivotsdkonitsdiagonal.RecoverAbyinvertingthosecolumnoperations(addinsteadofsubtract).Thisusesuppertriangularmatricesmultiplyingontheright,soA=LcUc.MovingthepivotmatrixD=diag(d1;:::;dn)fromLctoUcmustreproduceA=LcD 1DUc=LUasfoundbyrowoperations,becausethosefactorsareunique.IfanysubmatricesAkaresingular,A=LUisimpossible.ApermutationmatrixPisneeded.Innumericallinearalgebra(whereadditionalrowexchangesbringlargerentriesintothepivotpositions)itisusualtoimagineallexchangesdonerst.ThenthereorderedmatrixPAfactorsintoLU.Inalgebra(wherethesizeofthepivotisnotimportant)wekeeptherowsinplace.Eliminationisstillexecutedbyalowertriangularmatrix.Buttheoutcomemaynotbeuppertriangular,untilwereordertherowsbyfactoringoutP:2402510400335=24010100001352410502400335=PU:Theeliminationsteps(whichproducedthosezerosbelowtheentries1and2)areinvertedbyL.ThentheoriginalAisLPU.ToseethatPisunique,consideranyupperleftsubmatrixaofA:(3.1)a=`0pu0givesa=`pu:Ifahassrowsandtcolumns,then`issbysanduistbyt|bothwithnonzerodiagonalsandbothinvertible!Thereforethesbytsubmatrixphasthesamerankasa.SincetheranksofallupperleftsubmatricesparedeterminedbyA,thewholepermutationPisuniquelydeterminedinA=LPU[5;6;8].Thissimplestepisall-important.The1'sinPindicatepivotsinA.Thisoccursinthei;jpositionwhentherankoftheibyjupperleftsubmatrixaijjumpsabovetherankofai 1;jandai;j 1.(Byconventionai0anda0jhaverankzero.)Againthiscriteriontreatsrowsandcolumnsequally.EliminationbyroworbycolumnoperationsleadstothesameP. 6GILBERTSTRANGA2by2exampleshowsthatthetriangularLandUarenolongerunique:011a=10b101101c01provideda=b+c:Ifeliminationisbyrowoperations,wewillchooseb=atoclearoutcolumn2andreachc=0.Ifeliminationisbycolumnoperations,wewillchoosec=atoclearoutrow2andreachb=0.TheseparticularchoicesofLandUare\reducedontheright"and\reducedontheleft."ThematricesPUP 1andP 1LPareupperandlowertriangularrespectively(bothareidentitymatricesinthisexample).Foreachofthesereducedfactorizations|thenormalchoiceswhenconstructedbyelimination|allthreefactorsL,P,UareuniqueuptothediagonalpivotmatrixD.IfweincludeDasafourthfactorinA=LPDU,withdiagonal1'sinLandU,thenallreducedfactorsareunique.Thoseparagraphssummarizedtheknownalgebraofelimination\withPinthemiddle."Wewanttoaddonetrivialobservation.ItispromptedbytheBruhatfactorizationA=U1U2withtwouppertriangularfactors.ThepermutationwillnowbethereverseidentitymatrixJforagenericmatrixA.NoticethatU1isnotlowertriangular,andBruhatinthisgenericcaseisnotthesameasA=LU.OurobservationisthatfourinequivalentfactorizationsofAcomefromfourdierentstartingpointsforelimination.ThosestartingpointsarethefourcornerentriesofA.Weindicatetheshapesofthetriangularfactorsinthefourgenericcases,whentheeliminationsproceedwithoutmeetingzerosinthenaturalpivotpositions(ifazerodoesappear,thepermutationwillchangefromIorJ): an1anna1na11 (upandright)A=U1JU2 (downandright)A=LU A=UL(upandleft) A=L1JL2(downandleft)ThusBruhatcomesfromeliminatingstartingatan1.Whencolumn1isreducedtozeroabovethispivot,thenextpivotpositionis(n 1;2).Withnorowexchanges,upwardeliminationwillreacha\southeast"matrix.ThisbecomesuppertriangularwhenitsrowsarereorderedbyJ:A !26643775=26641111377526643775=JU2:ThestepsofupwardeliminationareinvertedbyanuppertriangularmatrixU1thatbringsbackA=U1JU2.Ifapivotentryiszero,thereverseidentityJchangestoadierentpermutation.Thesepermutationscomeinanaturalpartialorder(theBruhatorder)basedonthenumberoftranspositionsofneighbors.J=(n;:::;1)comesrstwiththemaximumnumberoftranspositions. BANDEDMATRICESWITHBANDEDINVERSESANDA=LPU74.InniteMatricesAissinglyinniteiftheindicesi;jarenaturalnumbers(i;jinN),anddoublyinniteifallintegersareallowed(i;jinZ).Wementionthreedicultieswiththefactorizationofinnitematrices.I.Forsinglyinnitematrices,eliminationstartswitha11.Eveniftherearenorowexchanges,thefactorsLandUcanbeunbounded.Thepivotscanapproach0and1.ConsiderablockdiagonalmatrixAwith2by2blocksBn:Bn="n110B 1n=011 "n"n!0:BnandB 1nstayboundedbuttheblocksinLandUwillgrowasn!1:Bn=LnUn=10" 1n1"n10 " 1n:ThusLandUareunbounded.II.Fordoublyinnitematrices,eliminationhasnonaturalstartingpoint.Insteadofarecursivealgorithm,weneedtodescribethedecisionsatstepkintermsoftheoriginalmatrixA.Hereisareasonableformulationofthatstep:ForeachkinZ,removeallcolumnsofAaftercolumnktocreateasubmatrixA(k)endingatcolumnk.DeneI(k)asthesetofallnumbersiinZsuchthatrowiofA(k)isnotalinearcombinationofpreviousrowsofA(k).ThesetI(k)hastheseproperties:(1)I(k)containsnonumbersgreaterthank+w.BythebandednessofA,therowsbeyondrowk+warezeroinA(k).(2)I(k)containseverynumberik w.AllthenonzerosinrowiofAarealsoinrowiofA(k),bybandedness.SinceAisinvertible,thatrowicannotbeacombinationofpreviousrows.(3)I(k)containsI(k 1).IfiisinI(k 1),thenrowiofA(k 1)isnotacombinationofpreviousrowsofA(k 1);sorowiofA(k)isnotacombinationofpreviousrowsofA(k).(4)IfmultiplesofpreviousrowsofA(k)aresubtractedfromlaterrowstoformamatrixB(k),thesetsI(k)arethesameforA(k)andB(k).Lemma4.1.I(k)containsexactlyonerownumberthatisnotinI(k 1).Callthatnewnumberi(k).EveryiinZisi(k)foronecolumnnumberk.Reasoning:ForeachithatisnotinI(k 1),rowiofA(k 1)isacombinationofpreviousrowsofA(k 1).SubtractfromeachofthoserowsiofA(k)thatcombinationofpreviousrowsofA(k).ThentheserowsiofthenewmatrixB(k)(formedfromA(k))areallzeroexceptpossiblyinitslastcolumnk.WemustshowthatexactlyoneoftheserowsofB(k)endsinanonzero.Thenitsrownumberi(notinI(k 1))isinI(k).ThepermutationmatrixPwillhaveaoneinthatrowi(k),columnk. 8GILBERTSTRANGSupposeI(k)containstworownumbersi1i2thatarenotinI(k 1).Thenrowsi1andi2ofB(k)haveonlyzeroentriesbeforecolumnk.Thereforerowi2isamultipleofrowi1.Thusi2cannotbelongtoI(k).SupposeI(k)containsnonewrownumbers,andequalsI(k 1).IfiisnotinI(k 1),eliminationcanproducezerosinrowiuptoandincludingcolumnk.Thentheideaistousecolumnoperationstoproducezerosinalltheremainingentriesofcolumnk.Thatisnowacolumnofzeros,whichcontradictstheinvertibilityoftheoriginalmatrixA.Thosecolumn(androw)operationswillproducezerosbyusingthepivotsalreadylocatedinpositions(i(j);j)forjk.YinghuiWangandIhavediscussedthissequenceofsteps.Fornitematricesnoadditionalhypothesiswillbeneeded.Butinnitematricesfollowtheirownrules,anditistooearlytogivesucientconditionsforA=LPUtobeachieved.III.Thereadermightenjoyastrikingexampleofthisthirddicultywithinnitematrices.ThesematriceshaveAB=IbutBx=0.Thus(AB)xisdierentfromA(Bx):(4.1)A=2666641111011100110001377775B=2666641 10001 10001 10001377775x=2666641111377775Theassociativelaw(AB)x=A(Bx)hasfailed!ThesumsanddierencesinAandBcorrespondtointegralsandderivatives(andalsoBA=I).RienKaashoekandRichardDudleyshowedussimilarexamples,andAlanEdelmanpointedoutthedisturbinganalogywiththefundamentaltheoremofcalculus.Theusualproofoftheassociativelawisarearrangementofadoublesum.Forinniteseries,thatrearrangementispermittedwhenthereisabsoluteconvergence.(Changingevery 1inBto+1willproducedivergenceinABx.)Moregenerally,(AB)x=A(Bx)forboundedoperatorsonaBanachspace.OurproblemistostaywithinthisframeworkwhenLandUcanbeunbounded.Noticetherelevanceofassociativityinourattemptedproofabove.Rowoper-ationsreducedA(k)toB(k),andthenrowandcolumnoperationsreducedcolumnktozero.DoesthissafelycontradicttheinvertibilityofA?(SectionsofinnitematricesareanalyzedbyLindnerin[9]|abeautifultheoryisdeveloping.)5.BandedPermutationsandtheShiftingNumberFactoringbandedpermutationsisacombinatorialproblemandGretaPanovashowedhowa\hookedwiringdiagram"yieldsP=F1:::FNwithN2wfactors.Inthenitecase[11],eachfactorFexecutesdisjointexchangesofneighbors.Theintersectionsofwiresindicatewhichneighborstoexchange.AsecondproofofN2wisgivenin[1].Thediagramhaswiresfrom1;2;3;4to3;4;1;2.ThisPhasw=2,eachFhasw=1,andN=3factorsarerequired.Theyweredisplayedinequation(4).Thedistancefromlefttorightis2w,andallhookedlineshaveslope 1=+1. BANDEDMATRICESWITHBANDEDINVERSESANDA=LPU9 i=1 2 34 y 1=p(3)2 3 4 F2 F1 F3 13 24 23 14InthisexampleF3yields1;3;2;4byonetransposition.ThetwoexchangesinF2yield3;1;4;2.ThenF1produces3;4;1;2.Threelinescannotmeetatthesamepoint,becausetwowouldbegoinginthesamedirection.Wemustprovethatintersectionsofhookedlinesoccuronatmost2w 1verticals.Supposethatijbutp(i)-474;.901;p(j).Thelinethroughtheleftpointx=0;y=iisy=i+x(slope+1becauseyincreasesdownward).Thelinethroughtherightpointx=2w;y=p(j)isy=p(j) x+2w.Thoselinesmeet(betweentheirhooks)atx=w+1 2(p(j) i).Weneedtoshowthatthereareonly2w 1possiblevaluesfortheintegerp(j) i.Thentherewillbeonly2w 1possiblevaluesforx,andthose2w 1verticallineswillincludealltheintersections.Bandednessgivesp(j) j w.Addingj i0(whichbecomesj i1forintegers)leavesp(j) i1 w.Thisisthedesiredboundononeside.Intheoppositedirectioni p(i) w.Addingp(i) p(j)1leavesi p(j)1 w.Sotheonlypossibilitiesfori p(j)arethe2w 1numbers1 w;:::;w 1.TheintersectinglinesrevealtheorderforthetranspositionsFiofneighbors,whoseproductisP.See[1;13]foragreedysequenceoftranspositionsFi.ThisfactorizationwithN2wextendstobandedsinglyinnitepermutations.Turnnowtodoublyinnitepermutations(ofZ).Anewpossibilityappears,becausetheleftshiftmatrixSisalsoapermutationwithbandwidthw=1(soSandtherightshiftST=S 1becomeadmissiblefactorsFiofP):S(:::;x0;x1;:::)=(:::;x1;x2;:::)hasSij=1onthesuperdiagonalj=i+1:Theorem5.1.AbandedpermutationPofZfactorsintoP=SsF1:::FNwithN2wandjsjw.Theshiftingindexs(P)(positiveornegative)hasthepropertythat(5.1)s(P1P2)=s(P1)+s(P2):Proof.ThepureshiftsP=SwandP=S wareextremecases.Forthefactorizationingeneral,werstuntanglethehookedwiresbyasequenceoftrans-positionsasbefore.Afteruntangling,thediagramwillshowashiftbys.OurexampleisapermutationPthathasperiodB=4andbandwidthw=2:p(4n+1)=4n+3;p(4n+2)=4n;p(4n+3)=4n+1;p(4n+4)=4n+2 10GILBERTSTRANGWedrawoneperiodofthewiringdiagramforP: 4 3 2 i=1 3 2 1 0=p(2) Threeconsecutivetranspositionswilluntanglethewires.Butourexamplestillhasashift:(1;2;3;4)!(0;1;2;3)aftertheuntangling.ThereforeP=SF1F2F3.FourrowsofthematrixforPwillshowoneperiodwithbandwidthw=2:(5.2)Pincludes26666400100000000100000000100001000000377775=M0M1:Thispermutationhasshiftingindexs=1.EverypermutationfactorsinthesamewayintoP=SsF1:::FN,untanglingfollowedbypossibleshiftsleftorright.AllfactorsarepermutationsofZwithbandwidthw=1.Therulefors(P1P2)comesfromthisfactorization.EachproductFSsisthesameasSsf,wherefisconstructedbymovingallthe2by2(and1by1)blockssplacesalongthediagonalofF.ShiftsinP2canthencombinewithshiftsinP1:(5.3)P1P2=(Ss1F1:::FN)(Ss2FN+1:::FM)=Ss1+s2f1fNFN+1:::FM:ThustheindexforP1P2iss1+s2=s(P1)+s(P2).TheshiftingindexforadoublyinniteinvertiblematriximitatestheFredholmindexforasinglyinnitematrix.ThatindexisdenedwhenthekernelsofAandAarenite-dimensional:(5.4)index(A)=dim(kernelofA) dim(kernelofA)TheindexofA1A2isthesumoftheseparateindices.Thusindex(A)=index(P)ifA=LPUwithinvertibleLandU.Similarlys(A)=s(P)intheinvertibledoublyinnitecase.ThereisaniceconnectionbetweentheFredholmindexandtheshiftingindex.Ifwestaywithpermutations,wecansketchasimpleproofofthisconnection:Theorem5.2.TheshiftingindexofabandeddoublyinnitepermutationequalstheFredholmindexofeverysinglyinnitesubmatrixPn(containingallentriesPijwithinandjn). BANDEDMATRICESWITHBANDEDINVERSESANDA=LPU11Proof.Forpermutations,theFredholmindexofPnisjustthenumberofzerocolumnsminusthenumberofzerorows(bothniteforbandedP).Nowremovearowandcolumn(vectorsrandc)toformPn+1.Ifr=c=(0;0;:::)orifr=c=(1;0;0;:::)thisindexisunchanged.Supposer=(0;0;:::)butccontainsa1fromsomerowinofPn.ThenthezerorowrwasremovedbutanewzerorowihasbeencreatedinPn+1.Theindexisagainunchanged(andsimilarlyifciszeroandrisnonzero).Whenbothcandrhave1's,theirremovalcreatesazerorowandazerocolumn.SotheindexofPnisindependentofn.ForthedoublyinniteshiftSs,allsinglyinnitesections(Ss)nhaveFredholmindexs.Fors0,allsectionsstartwithszerocolumnsandhavenozerorows.Fors0,theystartwith szerorowsandhavenozerocolumns.Tocompletetheproofforanybandedpermutations,weexpressitsfactorizationintheformP=f1:::fNSsandshowthattheFredholmindexofeveryPnstaysats(theshiftingindexofP).Theproofcanuseinduction.Whenfkexchangesrowsnandn+1ofthepermutationQ=fk+1:::fNSs,itwillalsoexchangethoserowsofthesinglyinnitesectionQn.TheFredholmindexofQnisunchanged.Fromtherststepinthisproofweconcludethatallexchangesofneighbors,fromeachfactorfk,leavetheindexofeverysectionunchanged.Soallthoseindicesstayats=s(P).Afterformulatingthistheoremonthetwoindices,welearnedfromMarkoLindnerthatitholdsforamuchwiderclassofdoublyinnitematrices[see9;14;15].Theoriginalproof[14]isverymuchdeeper,usingK-theory.Ourshiftingindexsisthe\plus-index"inthatliterature,recentandgrowingandimpressive.NoteTocomputes(P)fromourdenitionrequiresthefactorizationP=SsF1:::FN.Amoreintrinsicdenition(iftrue)comesfromtheaverageshiftfromitop(i):(5.5)Shiftingindexs(P)=limT!1 1 2T+1TX T(i p(i))!:Thispaperendswithasummaryoftheperiodic(blockToeplitz)case,forwhichallinformationaboutAiscontainedinthematrixpolynomialM(z).ThetriangularfactorizationofM(z)isalong-studiedandbeautifulproblem.Thediscussionofthisperiodiccasecouldextendtomatricesthatarenotbanded,butwedon'tgothere.6.PeriodicMatrices(BlockToeplitz)AsinglyordoublyinnitematrixhasperiodBifA(i+B;j+B)=A(i;j)fori;jinNori;jinZ:Rows1toBcontainasequence:::;M 1;M0;M1;:::ofBbyBblocks.Inthedoublyinnitecase,thosematricesarerepeatedupanddownthe\blockdiagonals" 12GILBERTSTRANGofA:A=266666664M0M1M2:::M 1M0M1M2:::M 2M 1M0M1M2::::::377777775:AsinglyinniteperiodicmatrixstartswithM0intherstblockasshown.Theblocksaboveandtotheleftarenotpresent.Periodicmatricesare\Toeplitz"or\stationary"or\lineartime-invariant"byblocks.ThenaturalapproachtotheiranalysisisthroughtheBbyBmatrixfunctionM(z)=XMjzj(thesymbolorthefrequencyresponseofA):Thematrixmultiplicationy=AxbecomesablockmultiplicationY(z)=M(z)X(z)whenweseparatethecomponentsofxandyintoblocksxiandykoflengthB:(6.1)Y(z)=Xykzk=XMjzjXxizi=M(z)X(z):Thisconvolutionruleistheessentialpieceofalgebraatthefoundationofdigitalsignalprocessing.Themapx!X(z)istheDiscreteTimeFourierTransform(inblocks).Inthisdoublyinnitecase,wemaymultiplyy=AxandtransformtogetY(z),orwemaytransformrstandmultiplyM(z)X(z).ThusFA=MF.Thesinglyinnitecasehasi0inPxizi,andweprojecty=Axtohavek0inPykzk.Thetwocasesaredierent,butthesymbolM(z)governsboth:Doublyinnite(1)AisbandedifM(z)hasnitelymanyterms(apolynomialinzandz 1).(2)AisalsoinvertibleifM(z)isinvertibleforeveryjzj=1.(3)A 1isrepresentedby(M(z)) 1whichinvolvesadivisionbydetM.SoA 1isalsobandedifdetM(z)isamonomialczm,c=0.(4)AisapermutationPifM(z)=D(z)pisadiagonalmatrixdiag(zk1;:::;zkB)timesaBbyBpermutationmatrixp.Thenpij=1correspondstoPiJ=1whenJ=j+kiB(equalindicesmodB).(5)A=LPUifM(z)=L(z)P(z)U(z).(6)TheshiftingindexofP(andA)isthesumofpartialindicess=Pki.AkeypointforusisthatthefactorizationintoL(z)P(z)U(z)hasbeenachieved.ThistheoremhasalonganddistinguishedhistorybeginningwithPlemelj[12].(G.D.Birkho'sfactorizationcorrespondstoPLU.)Ashortdirectproof,andmuchmore,isinthevaluableoverview[7].NoticethatanindependentproofofA=LPUbyeliminationoninnitematriceswouldprovideanewapproachtotheclassicalproblemoffactoringM(z)intoL(z)P(z)U(z).Thereareimportantchangesin1{6forsinglyinnitematricesA;L;P;U.Thosearestillperiodic(blockToeplitz).ButLUisnotperiodic;its1;1blockisL0U0butthe2;2blockincludesL 1U1.Thecorrectorderfortheseblocktriangular BANDEDMATRICESWITHBANDEDINVERSESANDA=LPU13matricesisUL.ThisistheWiener-Hopffactorizationthatsolvessinglyinniteperiodicsystems.A=ULisnotachievedbyelimination(whichwouldhavetostartatanonexistentlowerrightcornerofA),butitfollowsfromM(z)=U(z)L(z).Weindicatethechangesin(1){(6)forthesinglyinnitecase.NoticeespeciallythattheshiftingindexsbecomestheFredholmindexin(6).Butindexzeroisnotthesameasinvertibility.Sothosepropertiesareconsideredseparately.(1)UisbandedwhenU(z)isamatrixpolynomialinz.(2)UisinvertibleifU(z)isinvertibleforjzj1.IfUisbidiagonal,withthenumbersu0andu1ondiagonals0and1,weneedju0jju1j.(3)U 1isrepresentedby(U(z)) 1.U 1isbandedifdetU(z)isanonzeroconstant.(4)Asinglyinniteperiodicpermutation(invertible!)isblockdiagonal.(5)A=ULifM(z)=U(z)L(z).ThisisWiener-HopfwithP(z)includedinL(z).(6)IfP(z)=D(z)pwithD(z)=diag(zk1;:::;zkB)timesapermutationp,thentheFredholmindexofthematrixP(andofA=LPU)isPki.Theexampleinsection5(withperiodB=4)illustratestheFredholmindexinthesinglyinnitecase(6):P=264M0M100M0M100375M0=2666640010000100000100377775M1=2666640000000010000000377775Inthiscasedet(M0+M1z)=z.TheFredholmindexofPis1.ThekernelofPisspannedby(1;0;0;:::).ThediagonalmatrixD(z)isdiag(1;1;z;1).Themul-tiplicativepropertyofdet(P1(z)P2(z))conrmsthatindex(P1P2)=index(P1)+index(P2).ThoseindicesaretheexponentsofzinthedeterminantsofP1(z)andP2(z).AcknowledgementsManyfriendshavediscussedsectionsofthispaperwithme.Itisapleasuretothankeachofthemforgeneroushelp.FactoringCMVmatrices(VadimOlshevskyandPavelZhlobich)Factoringmatrixpolynomials(IlyaSpitkovsky,RienKaashoek,andPatrickDewilde)Factoringbandedpermutations(GretaPanova,AlexPostnikov,andChi-KwongLi)Bruhatfactorizations(DavidVoganandGeorgeLusztig)ShiftingindexandFredholmindex(MarkoLindner)Factoringinnitematrices(YinghuiWang) 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